Answer :
Let's work through the problem step-by-step to determine whether the events [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are independent.
### Definitions and Given Data
- Event [tex]\(A\)[/tex]: The item pulled out of the recycling bin is a plastic bottle.
- Event [tex]\(B\)[/tex]: A tenth grader recycled the item.
Using the table, we have:
- Total number of recycled items: [tex]\(400\)[/tex]
- Total number of plastic bottles: [tex]\(135\)[/tex]
- Total number of items recycled by tenth graders: [tex]\(150\)[/tex]
- Number of plastic bottles recycled by tenth graders: [tex]\(40\)[/tex]
### Calculating Probabilities
1. Probability of [tex]\(A\)[/tex]: Probability that the item is a plastic bottle.
[tex]\[ P(A) = \frac{\text{Number of plastic bottles}}{\text{Total number of recycled items}} = \frac{135}{400} \][/tex]
2. Probability of [tex]\(B\)[/tex]: Probability that a tenth grader recycled the item.
[tex]\[ P(B) = \frac{\text{Number of items recycled by tenth graders}}{\text{Total number of recycled items}} = \frac{150}{400} \][/tex]
3. Probability of both [tex]\(A\)[/tex] and [tex]\(B\)[/tex]: Probability that the item is a plastic bottle recycled by a tenth grader.
[tex]\[ P(A \cap B) = \frac{\text{Number of plastic bottles recycled by tenth graders}}{\text{Total number of recycled items}} = \frac{40}{400} \][/tex]
4. Conditional Probability [tex]\(P(A \mid B)\)[/tex]: Probability that, given the item was recycled by a tenth grader, it is a plastic bottle.
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{40}{400}}{\frac{150}{400}} = \frac{40}{150} \][/tex]
### Checking for Independence
The events [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are independent if either of the following conditions hold:
- [tex]\(P(A \cap B) = P(A) \cdot P(B)\)[/tex]
- [tex]\(P(A \mid B) = P(A)\)[/tex]
#### Checking [tex]\(P(A \cap B) = P(A) \cdot P(B)\)[/tex]
1. Calculate [tex]\(P(A) \cdot P(B)\)[/tex]:
[tex]\[ P(A) \cdot P(B) = \left(\frac{135}{400}\right) \cdot \left(\frac{150}{400}\right) = \frac{135 \cdot 150}{400 \cdot 400} \][/tex]
2. Compare it with [tex]\(P(A \cap B)\)[/tex]:
[tex]\[ P(A \cap B) = \frac{40}{400} \][/tex]
#### Checking [tex]\(P(A \mid B) = P(A)\)[/tex]
1. Calculate [tex]\(P(A \mid B)\)[/tex]:
[tex]\[ P(A \mid B) = \frac{40}{150} \][/tex]
2. Compare it with [tex]\(P(A)\)[/tex]:
[tex]\[ P(A) = \frac{135}{400} \][/tex]
After these comparisons, we determine that:
- [tex]\(P(A \mid B) \neq P(A)\)[/tex]
- [tex]\(P(A \cap B) \neq P(A) \cdot P(B)\)[/tex]
### Conclusion
Since neither condition for independence holds, we conclude:
- [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are not independent events because [tex]\(P(A \mid B) \neq P(A)\)[/tex].
Thus, the correct statement is:
[tex]$A$[/tex] and [tex]$B$[/tex] are not independent events because [tex]$P(A \mid B) \neq P(A)$[/tex].
### Definitions and Given Data
- Event [tex]\(A\)[/tex]: The item pulled out of the recycling bin is a plastic bottle.
- Event [tex]\(B\)[/tex]: A tenth grader recycled the item.
Using the table, we have:
- Total number of recycled items: [tex]\(400\)[/tex]
- Total number of plastic bottles: [tex]\(135\)[/tex]
- Total number of items recycled by tenth graders: [tex]\(150\)[/tex]
- Number of plastic bottles recycled by tenth graders: [tex]\(40\)[/tex]
### Calculating Probabilities
1. Probability of [tex]\(A\)[/tex]: Probability that the item is a plastic bottle.
[tex]\[ P(A) = \frac{\text{Number of plastic bottles}}{\text{Total number of recycled items}} = \frac{135}{400} \][/tex]
2. Probability of [tex]\(B\)[/tex]: Probability that a tenth grader recycled the item.
[tex]\[ P(B) = \frac{\text{Number of items recycled by tenth graders}}{\text{Total number of recycled items}} = \frac{150}{400} \][/tex]
3. Probability of both [tex]\(A\)[/tex] and [tex]\(B\)[/tex]: Probability that the item is a plastic bottle recycled by a tenth grader.
[tex]\[ P(A \cap B) = \frac{\text{Number of plastic bottles recycled by tenth graders}}{\text{Total number of recycled items}} = \frac{40}{400} \][/tex]
4. Conditional Probability [tex]\(P(A \mid B)\)[/tex]: Probability that, given the item was recycled by a tenth grader, it is a plastic bottle.
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{40}{400}}{\frac{150}{400}} = \frac{40}{150} \][/tex]
### Checking for Independence
The events [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are independent if either of the following conditions hold:
- [tex]\(P(A \cap B) = P(A) \cdot P(B)\)[/tex]
- [tex]\(P(A \mid B) = P(A)\)[/tex]
#### Checking [tex]\(P(A \cap B) = P(A) \cdot P(B)\)[/tex]
1. Calculate [tex]\(P(A) \cdot P(B)\)[/tex]:
[tex]\[ P(A) \cdot P(B) = \left(\frac{135}{400}\right) \cdot \left(\frac{150}{400}\right) = \frac{135 \cdot 150}{400 \cdot 400} \][/tex]
2. Compare it with [tex]\(P(A \cap B)\)[/tex]:
[tex]\[ P(A \cap B) = \frac{40}{400} \][/tex]
#### Checking [tex]\(P(A \mid B) = P(A)\)[/tex]
1. Calculate [tex]\(P(A \mid B)\)[/tex]:
[tex]\[ P(A \mid B) = \frac{40}{150} \][/tex]
2. Compare it with [tex]\(P(A)\)[/tex]:
[tex]\[ P(A) = \frac{135}{400} \][/tex]
After these comparisons, we determine that:
- [tex]\(P(A \mid B) \neq P(A)\)[/tex]
- [tex]\(P(A \cap B) \neq P(A) \cdot P(B)\)[/tex]
### Conclusion
Since neither condition for independence holds, we conclude:
- [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are not independent events because [tex]\(P(A \mid B) \neq P(A)\)[/tex].
Thus, the correct statement is:
[tex]$A$[/tex] and [tex]$B$[/tex] are not independent events because [tex]$P(A \mid B) \neq P(A)$[/tex].
